A small circular hole 6.00 mm in diameter is cut in the side of a large water tank, 14.0 m below the water level in the tank. The top of the tank is open to the air. Find (a) the speed of efflux of the water and (b) the volume discharged per second. ((a)16.6 m/s, (b) 468 cm³ s⁻¹) 



Given:

Diameter of circular hole = d = 6.00 mm = 0.006 m
Height = h = 14.0 m

To Find:

(a) the speed of efflux of the water = `\v_{eff}`= ? 
(b) the volume discharged per second = `\frac {ΔV}{Δt}` = ?

Solution:

(a) the speed of efflux of the water = `\v_{eff}` = ? 

The efflux speed can be calculated by Torricelli's Theorem

`\v_{eff}` = `\sqrt {2gh}`

by putting values 

`\v_{eff}`= `\sqrt {2 ｘ 9.8m s⁻² 14.0 m}`

`\v_{eff}`= `\sqrt {2825.28 m² s⁻²}`

`\v_{eff}` = 16.57 m s⁻¹ ----------Ans.


(b) the volume discharged per second = `\frac {ΔV}{Δt}` = ?

From the equation of continuity, the rate of volume flow is given by the relation

`\frac {ΔV}{Δt}` = A v

Here to find the Area (A) 

 A  = 𝜋 r² = 𝜋 (d/2)² = 𝜋 (0.006 m / 2)² = 2.827 ｘ 10⁻⁵ m²

`\frac {ΔV}{Δt}` =  2.827 ｘ 10⁻⁵ m² ｘ 16.57 m s⁻¹

`\frac {ΔV}{Δt}` =  468.4 ｘ 10⁻⁶ m³ s⁻¹

or

`\frac {ΔV}{Δt}` =  468.4 cm³ s⁻¹ ---------Ans.





************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK)     

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.